1. Describe the structure of the nucleosome and explain how it aids in the packaging of DNA. (p.65)
Nucleosomes are DNA plus protein octamers composed of two of each of the 4 core histones: H3, H4, H2A, and H2B. Nucleosomes wrap DNA around them "link beads on a string" or "string around tin cans" with linker histone H1 separating two neighboring nucleosomes. This coiling of the DNA helps package the DNA into a condensed structure. Six nucleosomes can wind up to form a solenoid, which can wind up to become a loop that winds up to become a miniband. Stacked minibands make up chromosomes.
2. What is Chargaff's rule and how does it account for the nature of duplex B-form DNA? (p.69)
Chargaff's rule is that in double stranded DNA, the number of A equals T and the number of G equals C. This is accounted for by complementry base-pairing that occurs between nucleotides in duplex B-form DNA.
3. Given the sequence of one strand of duplex B-form DNA, predict the sequence of the complementry strand. (p.71)
G=C, A=T (G equals C, A equals T)
Remember that complementry DNA strands are antiparallel so make sure the 5'-3' direction is correct.
i.e. the complementary sequence for 5'-TAGCGGT-3' is 5'-ACCGCTA-3' and not 5'-ATCGCCA-3'
4. What are the general topological properties of B-form DNA with respect to the location of the bases, the sugar, H-bonds between base and the polarity of complementry strands. (p.67, 68, 71)
Bases are attached to the 1' carbon in B-form DNA. Sugars are connected to each other in a 5'-3' phospho-sugar backbone with phosphodiester bonds connecting the 5' carbon with the 3' carbon. Three H-bonds connect between G and C bases and two H-bonds connect between A and T bases on complementary, antiparallel strands. B-form DNA is a right handed helix with a major and minor groove.
5. How do A, B and Z-form DNA differ? What are someother non-B-form DNA structures? (p.73, 75)
Z-form DNA is a left-handed double helix with its phosphate-sugar backbone arranged in a zigzag manner. B-form DNA is right handed DNA and the most commonly found form of DNA. A-form DNA is right-handed and formed at a higher hydration state and has more bases per turn than B-form; it resembles duplex RNA.
Tetraplex (G4) DNA can form from G-rich regions of DNA, forming a tetraplex with G-G-G Hoogsteen base pairing. The H-bonding pattern is different with different atoms participating in H-bonding and with one G H-bonding to two other bases. Disease can arise from defective helicases that fail to unwind G4 DNA structures.
There are also Triplex DNA, Parallel Strand DNA (both strands arranged with the same polarity with respect to each other), Hairpin structures (seen on ends of some viral DNAs), and Three-way and Four-way junctions (recombination).
6. Given a projection of B-form DNA, locate the major groove, the minor groove, the phosphodiester back bone and the axis of the helix. (p.77)
7. Be able to distinguish the structures of different bases and sugars present in DNA and RNA. (p.67)
8. Describe the nature of intercalation and how intercalating dyes and drugs interact with DNA. How does a pure minor groove-binding agent differ from a pure intercalator? What types of interactions with bases are involved in intercalation? (p.76, 77)
Intercalating agents typically have a planar ring structure that can fit between bases of DNA and cause some unwinding and lengthening of DNA. A pure minor groove-binding agent only associates itself with the minor groove of the DNA and doesn't intercalate into the bases, though some proteins like TATA binding protein can actually bind to and bend DNA. Intercalating often makes new H-bond contacts with base pairs and recognizes specific base-sequences.
9. Explain how interaction between bases contributes to duplex stability. To what extent is duplex stability determined by base stacking and what contribution does hydrogen bonding between bases make? (p.74, 93)
Hydrogen bonding between bases actually contributes less to duplex stability than base stacking. Stacking interactions are caused by hydrophobic interactions between planar bases and make the largest contribution to stability of duplex DNA. H-bonding serves to stabilize the helix.
10. How is duplex stability influenced by ionic strength and how is the effect of ionic strength explained? (p.93)
Ionic strength from higher concentrations of monovalent cation stabilizes DNA because the repulsive charge on the phosphate groups are neutralized.
11. How is duplex stability reflected in the melting temperature (Tm)? (p.92)
The Tm is the temperatuer when half of the DNA sample is single stranded. Higher duplex stability is reflected by higher Tm. Duplex stability is influenced by the base pair composition (GC vs. AT content), ionic strength (concentration on monovalent cations), and DNA sequence (neighboring bases affect base stacking).
12. How would Tm (melting temperature) of A-T rich DNA differ from that of G-C rich DNA and how can one account for this difference in terms of interactions between bases? (p.92-93)
Tm of a A-T rich DNA results in lower Tm compared to G-C rich DNA because G-C content provides more favorable base stacking and more favorable H-bonding between bases.
13. What are restriction endonucleases and what function do they serve in microorganisms? What is a palindromic sequence (p.103)
Restriction endonucleases are enzymes that recognize and cleave specific DNA sequences which are often (but not always) palindromic. In microorganisms, restriction endonucleases are used as a defense mechanisms to cut up foreign invading DNA; the microorganism protects its own DNA by methylating it.
Restriction enzymes may cleaveto leave staggered ends or blunt ends.
Palindromic sequences are sequences that read the same forwards and backwards.
14. Explain how restriction endonucleases are employed in identifying allelic differences in genes. (p.105)
Restriction endonucleases can cleave at different locations on different alleles, resulting in restriction fragment length polymorphisms (RFLPs) The restulting differences in the restriction map of a gene or chromosomal locus can be used to tell the different between two alleles or individuals based on fragment lengths.
15. Without giving every gory detail, explain what a southern blot is and how this technique is used in genetic analysis. How is it used to study polymorphisms in human populations? (p.106, 107, 109)
Southern blot is a method to identify specific DNA sequences from a DNA mixture. DNA fragments are generated using restriction endonucleases and separated by length via gel electrophoresis. After separation, the DNA is denatured by alkali and transfred to a piece of nitrocellulose filter paper by capillary action. The filterpaper is then mixed with a nuclei acid probe (DNA, RNA, or synthetic oligonucleotide) which will seach out its complement on the filter paper and anneal to it. Exposure of the filter to x-ray fil will produce an image of bands indicating where on the gel the probe is accumulated.
Human polymorphisms can be studied this way as different RFLPs will result in a different patternof bands on a southern blot. If a specific band pattern can be linked to the assortment of a disease or a trait, it can be traced in a family pedigree.
16. Given a set of southern blot data for three individuals, we bale to explain how the results are consistent or not consistent with their being related. (p.110-111)
Look for common banding patterns between the individuals. If a band can be explained by assortment from a parent, then those individuals can possibly be related to each other. If a band cannot be explained by assortment from a parent, then those individuals are probably not related to each other. Presence of common RLFP is not indicative of a relationship between two individuals -- the frequency of a certain RLFP appearing in the random population must also be assessed.
17. How does a processive enzyme differ from one that is not processive?
A processive enzyme is something like Taq polymerase that binds on the DNA and stays on it for a while to perform its function. A non-processive enzyme is like a restriction endonuclease that just gets on the sequence it recognizes, performs its function, and then falls off. It doesn't stay on the DNA for an extended time.
18. Explain the principles behind the dideoxy method for DNA sequencing. Why are dideoxy nucleotides employed? (p.155-158)
Dideoxy NDA nucleotides lack an OH at the 3' end so will prevent further addition of nucleotides after incorporation to a replicating DNA strand. These DNA chain terminators are also used in the treatment of HSV infections, acute lymphocytic leukemia, and AIDS.
19. Explain the distinction between conservative and semiconservative DNA replication. (p.162)
Semiconservative
Parental strands hybridize with newly synthesized strands. As replication continues, parental hybrids will always remain though the concentration of DNA with two completely new strands will increase.
Conservative
Parental strands always stay together throughout replication. Newly synthesized strands never contain parental strands and the two parental strands will always remain together.
20. Explain the principles behind the dideoxy enxymatic sequencing method. (p.155-158)
Take 4 sets of DNA with all 4 necessary nucleotides and allow synthesis. In each one, add a different dideoxynucleotide (i.e., either a ddTTP, ddCTP, ddGTP, or ddATP). As replication procedes, dideoxy nucleotides will be incorportated into the growing strand and terminate synthesis where ever the corresponding deoxynucleotide should have bound. That is, in the ddTTP set, the DNA synthesis will terminate whenever a dTTP is normally incorporated but a ddTTP is incorporated instead.
When the 4 sets of DNA are run on gel electrophoresis with a different column for each nucleotide, different restriction fragments will be separated based on when a specific DNA strand terminated. The specific sequence can be determined simply by reading the bands from each column of the gel from top to bottom.
T
C
G
A
Sequence
---
A
---
G
---
T
---
T
---
G
---
C
---
T
---
A
22. Explain the physical principles behind the experiments of Meselson and Stahl. How did their experiments establish that DNA is replicated in a semi-conservative fashion? (p.163-164)
The M-S experiment involved growing bacteria in heavy 15NH4Cl media so that all the nitrogen in its DNA will incorporate the heavy 15N. When transfered to the ligher 14NH4Cl media, the first round of DNA replication will produce hyrbid DNA with one strand being heavy 15N and the other being light 14N. Continued rounds of replication will produce more and more light/light 14N/14N DNA, but always retaining some heavy/light 15N/14N hydrid DNA. These results established that DNA is replicated in a semi-conservative fashion rather than a conservative fashion. If DNA was replicated conservatively, no heavy/light 15N/14N hybrid DNA would ever be made and 15N heavy/heavy DNA would always persist.
Because the heavy/heavy 15N/15N, heavy/light 15N/14N, and light/light 14N/14N DNA differ in density, they can be separated by simply centrifuging them in a deonsity gradient.
23. What is DNA primase and what is its function in DNA replication (p.168)
DNA primase is a specialized RNA polymerase that synthesizes an RNA primer that is used in DNA replication. It synthesizes a short stretch of RNA (up to 10 nucleotides) and dissociates.
24. How does primer synthesis in DNA replication differ between eukaryotic and prokaryotic organisms? (p.178)
Primer synthesis in DNA replication between eurkaryotes and prokaryotes is superficially similar but with different enzymes involved. Prokaryotes use Primase which makes a 10 nucleotide RNA primer. Eukaryotes uses Polymerase Alpha which adds a 10 nucleotide RNA segment and then a 10-20 nucleotide DNA segment as a primer.
25. What is the direction (polarity) of synthesis of all polynucleotide synthesizing enzymes? (p.168)
The direction of synthesis of all polynucleotide synthesizing enzymes is 5'-3'.
26. What is the function of the 3'-5' exonuclease activity associated with DNA polymerases III and I and why is this activity essential for the fidelity of DNA replication? (p.168, 170)
The 3'-5' exonuclease activity is used for proofreading by DNA pol II and I. This ability removes mismatched bases that occurs when the wrong nucleotide is inserted during synthesis. It is responsible for maintaining the high fidelity of DNA replication; lack of proofreading results in higher error rates which can be very bad for a cell.
27. What is the function of the 5'-3' exonuclease activity in DNA replication and which enzyme has this activity? (p.168)
5'-3' exonculease activity is necessary to actively remove the primer during DNA replication, function in DNA repair, and fills in gaps between lagging strands. DNA pol I has this activity.
28. Why is 3'-5' exonuclease not an essential activity for reverse transcriptases of retroviruses? What is a consequence for viral replication of reverse transcriptase lacking this activity? (p.170)
Retroviruses can cope with a high error rate possibly due to their smaller genome and higher replication rate. Infact, the low genomic fidelity of retroviruses helps contribute to its genetic diversity. The consequence of lacking a 3'-5' exonuclease activity is a much higher rate of error during replication.
29. Explain the function of DNA polymerase I in repair in E. coli. (p.172)
DNA pol I removes primers and fills in the gaps between okazaki fragments.
30. How are both leading and lagging strand synthesis made by a single multi-enzyme replication complex? (p.172, 173)
DNA polymerase binds to both leading and lagging strands at a replication fork, looping the lagging strand around so that synthesis of both strands occurs in the 5'-3' direction. The leading strand is made continuously using one primer, but the lagging strand is made a piece at a time using multiple primers.
31. Explain the disctinction between leading and lagging strand in DNA replication. (p.171)
The leading strand is made continuously using one primer, but the lagging strand is made a piece at a time using multiple primers. This is because DNA synthesis must occur in the 5'-3' direction and cannot be reversed in the lagging strand.
32. What is an Okazaki fragment and how are such segments of DNA dealt with during DNA replication? (p.172)
Okazaki fragments are short DNA fragments present briefly in the vacinity of the replicaiton fork. They are formed on the lagging strand which must synthesize DNA a piece at a time using multiple primers, leaving Okazaki fragments every 200 bp in eukaryotes (1,000 bp in prokaryotes). The primers are removed and the gaps between fragments filled in by DNA pol I in prokaryotes and DNA Pol Beta in eukaryotes.
33. Explain the difference between type I and type II topoisomerases with respect to how they act on DNA and their requirements for catalytic activity. How are these enzymes exploited as targets for drugs? (p.174)
Type I cuts one strand of the duplex to relieve supercoiling and reseals the nick; no ATP hydrolysis is required and supercoil is removed one at a time.
Type II cuts both strands of the DNA duplex so that one strand passes over the other and introduces ATP-dependent negative supercoiling; can catalyze catenation-decantenation reactions. Both types form tyrosyl DNA-protein covalent intermediates.
Both enzymes are targets for anti-cancer drugs and antibotics that stabilize the DNA-protein covalent intermediate, inducing double strand breaks. Examples include antineoplastic agents such as etopocide, and campothecins and quinolone antimicrobial agents.
34. Why is DNA gyrase essential for progression of the replication fork in E. coli? (p.174-176)
DNA gyrase (E. coli topoisomerase II) removes positive supercoils formed ahead of the replication fork by introducing negative supercoils to relax the DNA. DNA gyrase is required for viability of E. coli to prevent stalls in replication.
35. What are DNA helicases and how do they differ from topoisomerases? Why are helicases essential for replication? (p.176)
DNA helicases separate DNA strands. DNA strands must be separated to expose the template for DNA replication to occur. Topoisomerases stablizes DNA by reducing supercoiling torsional stresses.
36. What are some of the functions of telomeres? How are telomeres synthesized and how does their synthesis differ from synthesis of the rest of the chromosomal DNA? (p.182-185)
Because eurkaryoteslack a simple mechanism that ensures that replication will always start at the end of a chromosome, successive rounds of replication would results in deletions from the end of the chromosome. Telomersare specialized termini at the ends of chromosomes that act to maintain the length of chromosomes with each round of replication. This protects the ends of chromosomes from recombination, fusion, and recognition by damage repair systems.
Telomeres are synthesized independent of DNA replication by telomerases which has a bound RNA component complementry to the repeat element of telomere. Telomerase binds to the ends of chromosomes and uses its RNA component as a primer to extend DNA replication beyond the end of the chromosome to protect the ends from loss.
37. What is the connection between the expression of telomoerase and the process of cancer and aging? (p.185)
Without telomerase, the endsof chromosomes would not be maintained and be lost with each replication. As a result, eukaryotic cells without telomerase expression will undergo senescence after a set number of replication cycles before they just die. Germline cells and cells that require constant replacement typically express telomerase but most differentiated cells do not. Thus, skin fibroblasts from older people can undergo fewer divisions before senescence than skin fibroblasts from younger people.
Telomerase is also expressed in 85%-90% of human tumors which immortalizes cells and allows them to continue to divide. As a result, tolermase is a target for developing anti-neoplastic drugs.
38. Explain how thymine dimers that form as a result of exposure to UV light are removed and the DNA repaired? (p.186)
Thymidine dimers can form as a result of UV light by photolyase. Basically, photolyase (which is light activated) breaks the cyclobutane ring of the thymine dimer to repair the DNA.
39. What are DNA ligases and how do they function in DNA replication and repair? (p.188, 190)
Lyases recognize gaps/nicks in the DNA phosphoribose backbone and fills in the gap.
40. Describe the steps in base excision repair and nucleotide excision repair? (p.187-189)
Base Excision Repair
Repairs single bases by removing it using DNA glycoslyase which cleaves the damanged base from the deoxyribose sugar. Endonucleases then cut the phosphodiester backbone leaving a gap in one strand. DNA polymerase Beta (DNA polymerase I in E. coli) will then come and fill in the nucleotide gap by starting synthesis from the free 3'-OH, removing DNA and converting the nick into intact DNA. Lastly, DNA ligase seals the phosporibose backbone gap.
Nucleotide Excision Repair
Repairs damange to a DNA segment by unwinding the DNA segment and using an excision nuclease to cut the DNA strand at sites flaking the damaged region. The nuclease complex actually moves to the damanged region, partially denatures it, kinks the DNA using ATP, and excises a 12 bp segment surrounding the damaged region. The segment between nicks is removed by helicase and metabolized. DNA polymerase delta/epsilon (DNA polymerase I in E. coli) then comes and fills in the gap in the DNA. Lastly, DNA ligase seals the nick in the phosphoribose backbone.
Objectives
1. Describe the structure of the nucleosome and explain how it aids in the packaging of DNA. (p.65)
Nucleosomes are DNA plus protein octamers composed of two of each of the 4 core histones: H3, H4, H2A, and H2B. Nucleosomes wrap DNA around them "link beads on a string" or "string around tin cans" with linker histone H1 separating two neighboring nucleosomes. This coiling of the DNA helps package the DNA into a condensed structure. Six nucleosomes can wind up to form a solenoid, which can wind up to become a loop that winds up to become a miniband. Stacked minibands make up chromosomes.
2. What is Chargaff's rule and how does it account for the nature of duplex B-form DNA? (p.69)
Chargaff's rule is that in double stranded DNA, the number of A equals T and the number of G equals C. This is accounted for by complementry base-pairing that occurs between nucleotides in duplex B-form DNA.
3. Given the sequence of one strand of duplex B-form DNA, predict the sequence of the complementry strand. (p.71)
G=C, A=T (G equals C, A equals T)
Remember that complementry DNA strands are antiparallel so make sure the 5'-3' direction is correct.
i.e. the complementary sequence for 5'-TAGCGGT-3' is 5'-ACCGCTA-3' and not 5'-ATCGCCA-3'
4. What are the general topological properties of B-form DNA with respect to the location of the bases, the sugar, H-bonds between base and the polarity of complementry strands. (p.67, 68, 71)
Bases are attached to the 1' carbon in B-form DNA. Sugars are connected to each other in a 5'-3' phospho-sugar backbone with phosphodiester bonds connecting the 5' carbon with the 3' carbon. Three H-bonds connect between G and C bases and two H-bonds connect between A and T bases on complementary, antiparallel strands. B-form DNA is a right handed helix with a major and minor groove.
5. How do A, B and Z-form DNA differ? What are someother non-B-form DNA structures? (p.73, 75)
Z-form DNA is a left-handed double helix with its phosphate-sugar backbone arranged in a zigzag manner. B-form DNA is right handed DNA and the most commonly found form of DNA. A-form DNA is right-handed and formed at a higher hydration state and has more bases per turn than B-form; it resembles duplex RNA.
Tetraplex (G4) DNA can form from G-rich regions of DNA, forming a tetraplex with G-G-G Hoogsteen base pairing. The H-bonding pattern is different with different atoms participating in H-bonding and with one G H-bonding to two other bases. Disease can arise from defective helicases that fail to unwind G4 DNA structures.
There are also Triplex DNA, Parallel Strand DNA (both strands arranged with the same polarity with respect to each other), Hairpin structures (seen on ends of some viral DNAs), and Three-way and Four-way junctions (recombination).
6. Given a projection of B-form DNA, locate the major groove, the minor groove, the phosphodiester back bone and the axis of the helix. (p.77)
7. Be able to distinguish the structures of different bases and sugars present in DNA and RNA. (p.67)
Ribose
Deoxyribose
Adenine
Cytosine
Guanine
Thymine
Uracil
8. Describe the nature of intercalation and how intercalating dyes and drugs interact with DNA. How does a pure minor groove-binding agent differ from a pure intercalator? What types of interactions with bases are involved in intercalation? (p.76, 77)
Intercalating agents typically have a planar ring structure that can fit between bases of DNA and cause some unwinding and lengthening of DNA. A pure minor groove-binding agent only associates itself with the minor groove of the DNA and doesn't intercalate into the bases, though some proteins like TATA binding protein can actually bind to and bend DNA. Intercalating often makes new H-bond contacts with base pairs and recognizes specific base-sequences.
9. Explain how interaction between bases contributes to duplex stability. To what extent is duplex stability determined by base stacking and what contribution does hydrogen bonding between bases make? (p.74, 93)
Hydrogen bonding between bases actually contributes less to duplex stability than base stacking. Stacking interactions are caused by hydrophobic interactions between planar bases and make the largest contribution to stability of duplex DNA. H-bonding serves to stabilize the helix.
10. How is duplex stability influenced by ionic strength and how is the effect of ionic strength explained? (p.93)
Ionic strength from higher concentrations of monovalent cation stabilizes DNA because the repulsive charge on the phosphate groups are neutralized.
11. How is duplex stability reflected in the melting temperature (Tm)? (p.92)
The Tm is the temperatuer when half of the DNA sample is single stranded. Higher duplex stability is reflected by higher Tm. Duplex stability is influenced by the base pair composition (GC vs. AT content), ionic strength (concentration on monovalent cations), and DNA sequence (neighboring bases affect base stacking).
12. How would Tm (melting temperature) of A-T rich DNA differ from that of G-C rich DNA and how can one account for this difference in terms of interactions between bases? (p.92-93)
Tm of a A-T rich DNA results in lower Tm compared to G-C rich DNA because G-C content provides more favorable base stacking and more favorable H-bonding between bases.
13. What are restriction endonucleases and what function do they serve in microorganisms? What is a palindromic sequence (p.103)
Restriction endonucleases are enzymes that recognize and cleave specific DNA sequences which are often (but not always) palindromic. In microorganisms, restriction endonucleases are used as a defense mechanisms to cut up foreign invading DNA; the microorganism protects its own DNA by methylating it.
Restriction enzymes may cleaveto leave staggered ends or blunt ends.
Palindromic sequences are sequences that read the same forwards and backwards.
枯眼望遙山隔水，往來曾見幾心知。
壺空怕酌一杯酒，筆下難成和韻詩。
迷路阻人離別久，訊音無雁寄回遲。
孤燈夜守長寥寂，夫憶妻兮父憶兒。
14. Explain how restriction endonucleases are employed in identifying allelic differences in genes. (p.105)
Restriction endonucleases can cleave at different locations on different alleles, resulting in restriction fragment length polymorphisms (RFLPs) The restulting differences in the restriction map of a gene or chromosomal locus can be used to tell the different between two alleles or individuals based on fragment lengths.
15. Without giving every gory detail, explain what a southern blot is and how this technique is used in genetic analysis. How is it used to study polymorphisms in human populations? (p.106, 107, 109)
Southern blot is a method to identify specific DNA sequences from a DNA mixture. DNA fragments are generated using restriction endonucleases and separated by length via gel electrophoresis. After separation, the DNA is denatured by alkali and transfred to a piece of nitrocellulose filter paper by capillary action. The filterpaper is then mixed with a nuclei acid probe (DNA, RNA, or synthetic oligonucleotide) which will seach out its complement on the filter paper and anneal to it. Exposure of the filter to x-ray fil will produce an image of bands indicating where on the gel the probe is accumulated.
Human polymorphisms can be studied this way as different RFLPs will result in a different patternof bands on a southern blot. If a specific band pattern can be linked to the assortment of a disease or a trait, it can be traced in a family pedigree.
16. Given a set of southern blot data for three individuals, we bale to explain how the results are consistent or not consistent with their being related. (p.110-111)
Look for common banding patterns between the individuals. If a band can be explained by assortment from a parent, then those individuals can possibly be related to each other. If a band cannot be explained by assortment from a parent, then those individuals are probably not related to each other. Presence of common RLFP is not indicative of a relationship between two individuals -- the frequency of a certain RLFP appearing in the random population must also be assessed.
17. How does a processive enzyme differ from one that is not processive?
A processive enzyme is something like Taq polymerase that binds on the DNA and stays on it for a while to perform its function. A non-processive enzyme is like a restriction endonuclease that just gets on the sequence it recognizes, performs its function, and then falls off. It doesn't stay on the DNA for an extended time.
18. Explain the principles behind the dideoxy method for DNA sequencing. Why are dideoxy nucleotides employed? (p.155-158)
Dideoxy NDA nucleotides lack an OH at the 3' end so will prevent further addition of nucleotides after incorporation to a replicating DNA strand. These DNA chain terminators are also used in the treatment of HSV infections, acute lymphocytic leukemia, and AIDS.
19. Explain the distinction between conservative and semiconservative DNA replication. (p.162)
Semiconservative
Parental strands hybridize with newly synthesized strands. As replication continues, parental hybrids will always remain though the concentration of DNA with two completely new strands will increase.
Conservative
Parental strands always stay together throughout replication. Newly synthesized strands never contain parental strands and the two parental strands will always remain together.
20. Explain the principles behind the dideoxy enxymatic sequencing method. (p.155-158)
Take 4 sets of DNA with all 4 necessary nucleotides and allow synthesis. In each one, add a different dideoxynucleotide (i.e., either a ddTTP, ddCTP, ddGTP, or ddATP). As replication procedes, dideoxy nucleotides will be incorportated into the growing strand and terminate synthesis where ever the corresponding deoxynucleotide should have bound. That is, in the ddTTP set, the DNA synthesis will terminate whenever a dTTP is normally incorporated but a ddTTP is incorporated instead.
When the 4 sets of DNA are run on gel electrophoresis with a different column for each nucleotide, different restriction fragments will be separated based on when a specific DNA strand terminated. The specific sequence can be determined simply by reading the bands from each column of the gel from top to bottom.
22. Explain the physical principles behind the experiments of Meselson and Stahl. How did their experiments establish that DNA is replicated in a semi-conservative fashion? (p.163-164)
The M-S experiment involved growing bacteria in heavy 15NH4Cl media so that all the nitrogen in its DNA will incorporate the heavy 15N. When transfered to the ligher 14NH4Cl media, the first round of DNA replication will produce hyrbid DNA with one strand being heavy 15N and the other being light 14N. Continued rounds of replication will produce more and more light/light 14N/14N DNA, but always retaining some heavy/light 15N/14N hydrid DNA. These results established that DNA is replicated in a semi-conservative fashion rather than a conservative fashion. If DNA was replicated conservatively, no heavy/light 15N/14N hybrid DNA would ever be made and 15N heavy/heavy DNA would always persist.
Because the heavy/heavy 15N/15N, heavy/light 15N/14N, and light/light 14N/14N DNA differ in density, they can be separated by simply centrifuging them in a deonsity gradient.
23. What is DNA primase and what is its function in DNA replication (p.168)
DNA primase is a specialized RNA polymerase that synthesizes an RNA primer that is used in DNA replication. It synthesizes a short stretch of RNA (up to 10 nucleotides) and dissociates.
24. How does primer synthesis in DNA replication differ between eukaryotic and prokaryotic organisms? (p.178)
Primer synthesis in DNA replication between eurkaryotes and prokaryotes is superficially similar but with different enzymes involved. Prokaryotes use Primase which makes a 10 nucleotide RNA primer. Eukaryotes uses Polymerase Alpha which adds a 10 nucleotide RNA segment and then a 10-20 nucleotide DNA segment as a primer.
25. What is the direction (polarity) of synthesis of all polynucleotide synthesizing enzymes? (p.168)
The direction of synthesis of all polynucleotide synthesizing enzymes is 5'-3'.
26. What is the function of the 3'-5' exonuclease activity associated with DNA polymerases III and I and why is this activity essential for the fidelity of DNA replication? (p.168, 170)
The 3'-5' exonuclease activity is used for proofreading by DNA pol II and I. This ability removes mismatched bases that occurs when the wrong nucleotide is inserted during synthesis. It is responsible for maintaining the high fidelity of DNA replication; lack of proofreading results in higher error rates which can be very bad for a cell.
27. What is the function of the 5'-3' exonuclease activity in DNA replication and which enzyme has this activity? (p.168)
5'-3' exonculease activity is necessary to actively remove the primer during DNA replication, function in DNA repair, and fills in gaps between lagging strands. DNA pol I has this activity.
28. Why is 3'-5' exonuclease not an essential activity for reverse transcriptases of retroviruses? What is a consequence for viral replication of reverse transcriptase lacking this activity? (p.170)
Retroviruses can cope with a high error rate possibly due to their smaller genome and higher replication rate. Infact, the low genomic fidelity of retroviruses helps contribute to its genetic diversity. The consequence of lacking a 3'-5' exonuclease activity is a much higher rate of error during replication.
29. Explain the function of DNA polymerase I in repair in E. coli. (p.172)
DNA pol I removes primers and fills in the gaps between okazaki fragments.
30. How are both leading and lagging strand synthesis made by a single multi-enzyme replication complex? (p.172, 173)
DNA polymerase binds to both leading and lagging strands at a replication fork, looping the lagging strand around so that synthesis of both strands occurs in the 5'-3' direction. The leading strand is made continuously using one primer, but the lagging strand is made a piece at a time using multiple primers.
31. Explain the disctinction between leading and lagging strand in DNA replication. (p.171)
The leading strand is made continuously using one primer, but the lagging strand is made a piece at a time using multiple primers. This is because DNA synthesis must occur in the 5'-3' direction and cannot be reversed in the lagging strand.
32. What is an Okazaki fragment and how are such segments of DNA dealt with during DNA replication? (p.172)
Okazaki fragments are short DNA fragments present briefly in the vacinity of the replicaiton fork. They are formed on the lagging strand which must synthesize DNA a piece at a time using multiple primers, leaving Okazaki fragments every 200 bp in eukaryotes (1,000 bp in prokaryotes). The primers are removed and the gaps between fragments filled in by DNA pol I in prokaryotes and DNA Pol Beta in eukaryotes.
33. Explain the difference between type I and type II topoisomerases with respect to how they act on DNA and their requirements for catalytic activity. How are these enzymes exploited as targets for drugs? (p.174)
Type I cuts one strand of the duplex to relieve supercoiling and reseals the nick; no ATP hydrolysis is required and supercoil is removed one at a time.
Type II cuts both strands of the DNA duplex so that one strand passes over the other and introduces ATP-dependent negative supercoiling; can catalyze catenation-decantenation reactions. Both types form tyrosyl DNA-protein covalent intermediates.
Both enzymes are targets for anti-cancer drugs and antibotics that stabilize the DNA-protein covalent intermediate, inducing double strand breaks. Examples include antineoplastic agents such as etopocide, and campothecins and quinolone antimicrobial agents.
34. Why is DNA gyrase essential for progression of the replication fork in E. coli? (p.174-176)
DNA gyrase (E. coli topoisomerase II) removes positive supercoils formed ahead of the replication fork by introducing negative supercoils to relax the DNA. DNA gyrase is required for viability of E. coli to prevent stalls in replication.
35. What are DNA helicases and how do they differ from topoisomerases? Why are helicases essential for replication? (p.176)
DNA helicases separate DNA strands. DNA strands must be separated to expose the template for DNA replication to occur. Topoisomerases stablizes DNA by reducing supercoiling torsional stresses.
36. What are some of the functions of telomeres? How are telomeres synthesized and how does their synthesis differ from synthesis of the rest of the chromosomal DNA? (p.182-185)
Because eurkaryoteslack a simple mechanism that ensures that replication will always start at the end of a chromosome, successive rounds of replication would results in deletions from the end of the chromosome. Telomersare specialized termini at the ends of chromosomes that act to maintain the length of chromosomes with each round of replication. This protects the ends of chromosomes from recombination, fusion, and recognition by damage repair systems.
Telomeres are synthesized independent of DNA replication by telomerases which has a bound RNA component complementry to the repeat element of telomere. Telomerase binds to the ends of chromosomes and uses its RNA component as a primer to extend DNA replication beyond the end of the chromosome to protect the ends from loss.
37. What is the connection between the expression of telomoerase and the process of cancer and aging? (p.185)
Without telomerase, the endsof chromosomes would not be maintained and be lost with each replication. As a result, eukaryotic cells without telomerase expression will undergo senescence after a set number of replication cycles before they just die. Germline cells and cells that require constant replacement typically express telomerase but most differentiated cells do not. Thus, skin fibroblasts from older people can undergo fewer divisions before senescence than skin fibroblasts from younger people.
Telomerase is also expressed in 85%-90% of human tumors which immortalizes cells and allows them to continue to divide. As a result, tolermase is a target for developing anti-neoplastic drugs.
38. Explain how thymine dimers that form as a result of exposure to UV light are removed and the DNA repaired? (p.186)
Thymidine dimers can form as a result of UV light by photolyase. Basically, photolyase (which is light activated) breaks the cyclobutane ring of the thymine dimer to repair the DNA.
39. What are DNA ligases and how do they function in DNA replication and repair? (p.188, 190)
Lyases recognize gaps/nicks in the DNA phosphoribose backbone and fills in the gap.
40. Describe the steps in base excision repair and nucleotide excision repair? (p.187-189)
Base Excision Repair
Repairs single bases by removing it using DNA glycoslyase which cleaves the damanged base from the deoxyribose sugar. Endonucleases then cut the phosphodiester backbone leaving a gap in one strand. DNA polymerase Beta (DNA polymerase I in E. coli) will then come and fill in the nucleotide gap by starting synthesis from the free 3'-OH, removing DNA and converting the nick into intact DNA. Lastly, DNA ligase seals the phosporibose backbone gap.
Nucleotide Excision Repair
Repairs damange to a DNA segment by unwinding the DNA segment and using an excision nuclease to cut the DNA strand at sites flaking the damaged region. The nuclease complex actually moves to the damanged region, partially denatures it, kinks the DNA using ATP, and excises a 12 bp segment surrounding the damaged region. The segment between nicks is removed by helicase and metabolized. DNA polymerase delta/epsilon (DNA polymerase I in E. coli) then comes and fills in the gap in the DNA. Lastly, DNA ligase seals the nick in the phosphoribose backbone.